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Does anyone know why subtracting and solving for $x$ converts the repeating decimal to a fraction?

This is my understanding of the steps for conversion, specific questions follow.

$;$ is the decimal point. $$ x = l_0,...,l_p;a_1,...,a_j,\dot{r_0},...,\dot{r_k} $$ Shift non-repeating digits to whole part: $$ 10^jx = l_0,...,l_p, a_0,...,a_j;\dot{r_0},...,\dot{r_k} $$ Shift repeating sequence into whole part: $$ 10^{j+k}x = l_0,...,l_p,a_0,...,a_j,r_0,...,r_k;r_{k+1},...,r_{k+\infty} $$ $$ 10^{j+k}x - 10^jx = 10^jx(10^k-1) = ? $$ I'm unsure about how to arrive at the RHS of this step because of the prefix digits $l_0, ..., l_p$. Does anyone know? Also, how is the difference the fractional representation? $$ x = \frac {r_0,...,r_k} {10^{j+k} - 10^j} $$

I also reviewed this, this and this post, however still have the above questions.

Nick
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1 Answers1

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Let's first consider a number between $0$ and $1$ with a recurring decimal expansion. In this answer, let $d_{1}d_{2}d_{3}\cdots d_{n}$ be the decimal expansion, and an overline represents recurring decimal expansion.

Let $p = 0.\overline{d_{1}d_{2}d_{3}\cdots d_{n}}$ and $q = d_{1}d_{2}d_{3}\cdots d_{n}$. Then, we can represent this as

$$p = \frac{q}{10^{n}} + \frac{q}{10^{2n}} + \frac{q}{10^{3n}} + \cdots$$ $$p = \sum_{i = 1}^{\infty}\frac{q}{10^{in}}.$$

Notice that this is a geometric series, converging to

\begin{align*}p &= \frac{\frac{q}{10^{n}}}{1 - \frac{1}{10^{n}}} \\ p &= \frac{\frac{q}{10^{n}}(10^{n})}{10^{n} - 1}\\p &= \frac{q}{10^{n} - 1}.\end{align*}

Then, $p$ can just be simplified if it is still not in the lowest terms.

In your case, you have a value $a > 1$. This can be expressed as $a = [a] + \{a\}$ where $[a]$ and $\{a\}$ is the integer part and the fractional part, respectively. We just solved the fraction form of $\{a\}$, hence adding $[a]$ to $\{a\}$ is just a matter of simplifying fractions.

Does anyone know why subtracting and solving for $x$ converts the repeating decimal to a fraction?

To answer your question, this is just a background process of solving for a converging geometric series. The process I gave is the simplified form already.

soupless
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